Determination of the estimated steam flow. Selection rules. Ratios of the main physical and operational parameters Calculation of the amount of steam per steaming

Steam consumption for industrial consumers

To determine the enthalpy of steam in a steam collector, it is necessary to use the tables of thermodynamic properties of water and steam given in. Required reference materials are provided in appendix B of this manual. According to Table B1, which shows the specific volumes and enthalpies of dry saturated steam and water on the saturation curve for a certain pressure, the following are given:

Saturation temperature - t O C(column 2);

The enthalpy of water on the saturation curve - , kJ / kg (column 5),

Steam enthalpy on the saturation curve - , kJ/kg (column 6).

If it is necessary to determine the enthalpies of steam and water at a pressure whose value is between the values ​​​​given in the table, then it is necessary to interpolate between two adjacent values ​​\u200b\u200bof the quantities between which the desired value is located.

The enthalpy of steam in a steam collector is determined by the steam pressure in it () according to Table B.1. Applications B.

The enthalpy of condensate returned from production is determined by its temperature and condensate pressure according to Appendix A.

Amount of condensate returned from production

where is the return of condensate from production (given).

Steam consumption to cover the heating and ventilation load

The temperature of the heating steam condensate at the outlet of the surface heater is assumed to be 10-15 o C higher than the temperature of the heated medium at the inlet to this heater. In heater 8, network water is heated, which enters it from the return pipeline of the heating network with a temperature of 70 o C. Thus, we take the temperature of the heating steam condensate at the outlet of heater 8 to be 85 o C.

According to this temperature and pressure of the condensate, according to the table in Appendix A, we find the enthalpy of the condensate:

Steam consumption for hot water supply

Steam consumption for heating plant

Total steam consumption to cover industrial and housing and communal loads

Steam consumption for auxiliary needs of the boiler house is taken in the range of 15-30% of the external load, i.e. steam consumption to cover industrial and housing and communal loads. Steam supplied for own needs is used in the thermal scheme of the boiler house for heating additional and make-up water, as well as for their deaeration.

We accept steam consumption for own needs equal to 18%. Subsequently, this value is specified as a result of the calculation of the thermal scheme of the boiler house.

Steam consumption for own needs:

Steam losses in the thermal scheme of the boiler house are 2-3% of the external steam consumption, we accept 3%.

The amount of steam supplied through the steam manifold after the reduction-cooling unit:

When steam passes through narrowed sections, a throttling process occurs, accompanied by a decrease in pressure, temperature, and an increase in the volume and entropy of the steam. For the case of an adiabatic throttling process, the following condition is satisfied:

where: - steam enthalpy after throttling, - steam enthalpy before throttling.

Thus, the steam energy does not change during the throttling process. The temperature of saturated steam is equal to the saturation (boiling) temperature and is a direct function of pressure. Since steam pressure and saturation temperature are reduced during throttling, some superheating of the steam occurs. In order for the steam to remain saturated after the reduction-cooling plant, feed water is supplied to it.

Water consumption at the ROU is determined by the ratio:

Steam enthalpy at the outlet of the boiler is determined by the pressure in the boiler drum according to Table B.1. Applications B,

The enthalpy of steam in the steam collector was determined by us earlier, .

The feed water pressure is assumed to be 10% higher than the pressure in the boiler drum:

The enthalpy of feed water at and a pressure of 1.5 MPa is determined from the table in Appendix A,.

Full performance of the boiler room.

At enterprises, water vapor is used for technological, household and power purposes.

For technological purposes, deaf and live steam is used as a coolant. Live steam is used, for example, to boil raw materials in boilies or to heat and mix liquids by bubbling, to create excess pressure in autoclaves, and also to change the state of aggregation of a substance (evaporation or evaporation of a liquid, drying materials, etc.). Deaf steam is used in surface heat exchangers with steam heating. The pressure of the steam used in meat processing enterprises ranges from 0.15 to 1.2 MPa (1.5 ÷ 12 kg / cm 2).

For each technological operation using water vapor, its consumption is determined according to the data heat balance everyone thermal process. In this case, the data of material balances of product calculations are used. For periodic processes, the heat treatment time for each cycle is taken into account.

In each particular case thermal load apparatus (expended heat) can be determined from the heat balance of the process. For example, the heat spent on heating the product from the initial ( t m) to the final ( t j) temperature for the apparatus continuous action, is determined by formula 72:

Q = Gc (t k – t n)φ, (72)

where Q- heat spent on heating, J / s (W), i.e. thermal load of the device;

G

With – specific heat product at its average temperature, J/kg K;

t to, t n – initial and final temperature, °С;

φ - coefficient taking into account heat loss to the environment
Wednesday ( φ = 1.03÷1.05).

The heat capacity of the product is chosen either from known directories, or calculated according to the principle of additivity for multicomponent systems.

To change the state of aggregation of a substance (solidification, melting, evaporation, condensation), thermal energy is consumed, the amount of which is determined by formula 73:

where Q is the amount of heat, J/s (W);

G is the mass flow rate of the product, kg/s;

r is the heat of phase transition, J/kg.

Meaning r determined according to reference data, depending on the type of product and the type of phase transition of the substance. For example, the heat of fusion of ice is taken to be r 0 \u003d 335.2 10 3 J / kg, fat

r w = 134 10 3 J / kg. The heat of vaporization depends on the pressure in the working volume of the apparatus: r = f (P a). At atmospheric pressure r= 2259 10 3 J/kg.

For continuous devices, heat consumption is calculated per unit of time (J / s (W) - heat flow), and for batch devices - for a cycle of operation (J). To determine the heat consumption per shift (day), it is necessary to multiply the heat flow by the operating time of the device per shift, day or by the number of cycles of the device operation periodical action and the number of such devices.

The flow rate of saturated water vapor as a heat carrier under the condition of its complete condensation is determined by the equation:

where D- the amount of heating water vapor, kg (or flow rate, kg / s);

Q total - total heat consumption or heat load of a thermal device (kJ, kJ / s), determined from the equation of the heat balance of the device;

– enthalpy of dry saturated steam and condensate, J/kg;

r is the latent heat of vaporization, kJ/kg.

The consumption of live steam for mixing liquid products (bubbling) is taken at the rate of 0.25 kg / min per 1 m 2 of the cross section of the apparatus.

Steam consumption for economic and domestic needs under this heading, steam is used to heat water for showers, laundry, washing floors and equipment, and scalding equipment.

Steam consumption for scalding equipment and inventory is determined by its outflow from the pipe according to the flow equation:

(75)

where D w – steam consumption for scalding, kg/shift;

d– inner diameter of the hose (0.02÷0.03 m);

ω – speed of steam outflow from the pipe (25÷30 m/s);

ρ - vapor density, kg / m 3 (according to Vukalovich's tables ρ = f(ρ ));

τ – scalding time, h (0.3÷0.5 h).

If we take in the equation τ = 1 h, then the steam consumption is determined in kg/h.

The calculation of steam consumption for all items is summarized in table 8.3.

Table 8.3 - Steam consumption, kg

Specific consumption steam is calculated using formula 76.

3.2.2 Calculation of steam consumption for heating and ventilation

The calculation of heat costs for heating and ventilation is determined by the formula:

Q=q · V · (t pom – t calc ) · T year , kW/year, (3.11)

where q is the specific heat consumption for heating and ventilation of 1m 3 of the room at a temperature difference of 1 ° C, kW / (m 3 deg).

The average value of this value can be taken: for heating - 0.45 · 10 -3 kW / (m 3 .deg), for ventilation 0.9 · 10 -3 kW / (m 3 .deg).

V - the total volume of the premises of the site, excluding the volume of drying chambers, m 3;

t pom is the temperature in the room, assumed to be 20°С;

t calc - design temperature for heating and ventilation;

T year - the duration of the heating season is determined by the formula:

T year \u003d 24 * τ from, h,

where τ from is the duration of the heating season, days.

T year = 24 · 205 = 4920 hours

Q from = 0,45 · 10 -3 · 4456,872 · (20-(-26)) · 4920 = 453,9 · 10 3 kW/year.

Q vent = 0,09 · 10 -3 · 4456,872 · (20-(-12)) · 4920 = 63,15 · 10 3 kW/year.

Table 3.3 - Calculation of heat consumption for heating and ventilation

The calculation of the annual demand for steam for heating and ventilation is determined by the formula:

3.2.3 Calculation of heat (steam) consumption for domestic needs

The calculation of heat (steam) consumption for domestic needs is determined by the formula:

where q - steam consumption per 1 person per shift;

m is the number of people working on the busiest shift;

n is the number of shifts in the work of the section (it is advisable to take 2);

τ is the number of days of operation of the site per year.

3.2.4 Calculation of the total annual steam demand for technological and domestic needs, heating and ventilation

The calculation of the total annual steam demand for technological and domestic needs, heating and ventilation is determined by the formula:

D common = D academic year + D from + D life , t/year. (3.14)

D common \u003d 8.13 + 891.47 + 2.6 \u003d 902.2 tons / year.

The article contains a fragment of the table of saturated and superheated steam. With the help of this table, according to the value of steam pressure, the corresponding values ​​of the parameters of its state are determined.

Column 1: Steam pressure (p)

The table shows the absolute value of the steam pressure in bar. This fact must be kept in mind. When it comes to pressure, as a rule, they talk about the excess pressure that the pressure gauge shows. However, process engineers use absolute pressure in their calculations. In practice, this difference often leads to misunderstandings and usually backfires.

With the introduction of the SI system, it was accepted that only absolute pressure should be used in calculations. All pressure measuring instruments technological equipment(except barometers) basically show gauge pressure, we mean absolute pressure. Under normal atmospheric conditions (at sea level) is understood barometric pressure 1 bar. Gauge pressure is usually indicated in barg.

Column 2: Saturated steam temperature (ts)

In the table, along with the pressure, the corresponding saturated steam temperature is given. The temperature at the appropriate pressure determines the boiling point of water and thus the temperature of saturated steam. The temperature values ​​in this column also determine the condensing temperature of the steam.

At a pressure of 8 bar, the saturated steam temperature is 170°C. The condensate formed from steam at a pressure of 5 bar has a corresponding temperature of 152°C.

Column 3: Specific volume (v”)

The specific volume is given in m3/kg. As the vapor pressure increases, the specific volume decreases. At a pressure of 1 bar, the specific steam volume is 1.694 m3/kg. Or in other words, 1 dm3 (1 liter or 1 kg) of water during evaporation increases in volume by 1694 times compared to the original liquid state. At a pressure of 10 bar, the specific volume is 0.194 m3/kg, which is 194 times that of water. The specific volume value is used in calculating the diameters of steam and condensate pipelines.

Column 4: Specific Gravity (ρ=po)

Specific gravity (also called density) is given in kJ/kg. It shows how many kilograms of steam are contained in 1 m3 of volume. With increasing pressure specific gravity increases. At a pressure of 6 bar, steam with a volume of 1 m3 has a weight of 3.17 kg. At 10 bar - already 5.15 kg and at 25 bar - more than 12.5 kg.

Column 5: Enthalpy of saturation (h')

The enthalpy of boiling water is given in kJ/kg. The values ​​in this column show how much heat energy is needed to bring 1 kg of water at a certain pressure to a boiling state, or how much heat energy is contained in the condensate, which at the same pressure condenses from 1 kg of steam. At a pressure of 1 bar, the specific enthalpy of boiling water is 417.5 kJ/kg, at 10 bar it is 762.6 kJ/kg, and at 40 bar it is 1087 kJ/kg. With increasing steam pressure, the enthalpy of water increases, and its share in the total enthalpy of steam is constantly growing. This means that the higher the vapor pressure, the more thermal energy remains in the condensate.

Column 6: Total enthalpy (h”)

Enthalpy is given in kJ/kg. This column of the table shows the steam enthalpy values. The table shows that the enthalpy increases up to a pressure of 31 bar and decreases with a further increase in pressure. At a pressure of 25 bar, the enthalpy value is 2801 kJ/kg. For comparison, the enthalpy value at 75 bar is 2767 kJ/kg.

Column 7: Thermal energy of vaporization (condensation) (r)

The enthalpy of vaporization (condensation) is given in kJ/kg. This column gives the amount of thermal energy required to completely evaporate 1 kg of boiling water at the appropriate pressure. And vice versa - the amount of thermal energy that is released in the process of complete condensation of (saturated) steam at a certain pressure.

At 1 bar r = 2258 kJ/kg, at 12 bar r = 1984 kJ/kg and at 80 bar r = only 1443 kJ/kg. With increasing pressure, the amount of thermal energy of vaporization or condensation decreases.

Rule:

With increasing steam pressure, the amount of thermal energy required to completely evaporate boiling water decreases. And in the process of condensing saturated steam at the appropriate pressure, less thermal energy is released.